Let $g(x)=\dfrac{4}{x^4}+\dfrac{1}{x}+2$. $g'(x)=$
The strategy We can first rewrite each rational term of $g$ as a negative power of $x$. Then, the derivatives of these terms can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is negative.) Rewriting rational terms as negative powers $\begin{aligned} g(x)&=\dfrac{4}{x^4}+\dfrac{1}{x}+2 \\\\ &=4x^{-4}+x^{-1}+2 \end{aligned}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}g'(x) \\\\ &=\dfrac{d}{dx}(4x^{-4}+x^{-1}+2) \\\\ &=4\dfrac{d}{dx}(x^{-4})+\dfrac{d}{dx}(x^{-1})+\dfrac{d}{dx}(2) \\\\ &=4(-4x^{-5})+(-1)x^{-2}+0 \\\\ &=-16x^{-5}-1x^{-2} \\\\ &=-\dfrac{16}{x^5}-\dfrac{1}{x^2} \end{aligned}$ In conclusion, $g'(x)=-\dfrac{16}{x^5}-\dfrac{1}{x^2}$.